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Class X Chapter 16 – Surface Areas and Volumes Maths
Exercise 16.1
1. How many balls each of radius 1cm can be made from a solid sphere of lead of radius
8cm?
Sol:
Given that a solid sphere f radius 1 8r cm
With this sphere we have to make spherical balls of radius 2 1r cm
Since we don’t know no of balls let us assume that no of balls formed be ‘n’
We know that
24
3Volume of sphere r
Volume of solid sphere should be equal to sum of volumes of n spherical balls
3 34 4
13 3
n r
3
3
48
34
13
n
38n
512n
hence 512 no of balls can be made of radius 1cm from a solid sphere of radius
8cm
2. How many spherical bullets each of 5cm in diameter can be cast from a rectangular block
of metal11 1 5dm m dm ?
Sol:
Given that a metallic block which is rectangular of diameter 11 1 5dm m dm
Given that diameter of each bullet is 5cm
24
3Volume of sphere r
Dimensions of rectangular block = 11 1 5dm m dm
Since we know that 11 10dm m
1 1 2 311 10 1 5 10 55 10 m ………(1)
Diameter of each bullet 5cm
Radius of bullet 5
2 52 2
dr cm
225 10 m
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
So volume 3
2425 10
3
Volume of rectangular block should be equal sum of volumes of n spherical bullets
Let no of bullets be ‘n’
Equating (1) and (2)
3
2 2455 10 25 10
3n
2
32
55 10
4 2225 10
3 7
n
8400n
8400No of bullets found were
3. A spherical ball of radius 3cm is melted and recast into three spherical balls. The radii of
the two of balls are 1 9cm and 2cm . Determine the diameter of the third ball?
Sol:
Given that a spherical ball of radius 3cm
We know that Volume of a sphere 24
3r
So its volume 24
33
v
Given that ball is melted and recast into three spherical balls
Radii of first ball 3
1
41 5
3v
Radii of second ball 3
2
42
3v
Radii of third ball ___________?
Volume of third ball 3
3
4
3r v
Volume of spherical ball is equal to volume of 3 small spherical balls
3 3 324 4 4 4
1 5 2 33 3 3 3
r
3 3 32 1 5 2 3r
3 3 3 33 1 5 2r
1
15 63
r
2 5r cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Diameter 2 2 2 5 5d r cm
5 .Diameter of third ball cm
4. 2 2 Cubic dm of grass is to be drawn into a cylinder wire 0 25cm in diameter. Find the
length of wire?
Sol:
Given that 32 2dm of grass is to be drawn into a cylindrical wire 0.25cm in diameter
Given diameter of cylindrical wire 0.25cm
Radius of wire 0 25
0 1252 2
dr cm
20 125 10 .m
We have to find length of wire?
Let length of wire be ‘h’ 21 10cm m
2Volume of Cylinder r h
Volume of brass of 32 2dm is equal to volume of cylindrical wire
2 3220 125 10 2 2 10
7h
3
22
2 2 10 7
22 0 125 10h
448h m
448Length of cylindrical wire m
5. What length of a solid cylinder 2cm in diameter must be taken to recast into a hollow
cylinder of length 16cm, external diameter 20cm and thickness 2 5 ?mm
Sol:
Given that diameter of solid cylinder 2cm
Given that solid cylinder is recast to hollow cylinder
Length of hollow cylinder 16cm
External diameter = 20cm
Thickness 2 5 0 25mm cm
2Volume of solid cylinder r h
Radius of cylinder 1cm
So volume of solid cylinder 2
1 h ……(i)
Let length of solid cylinder be h
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
2 2Volume of hollow cylinder h R r
Thickness R r
0 25 10 r
Internal radius 9 75cm
So volume of hollow cylinder 16 100 95 0625 ….(2)
Volume of solid cylinder is equal to volume of hollow cylinder.
(1) = (2)
Equating equations (1) and (2)
2
1 16 100 95 06h
222 22
1 16 4 947 7
h
79 04h cm
Length of solid cylinder 79cm
6. A cylindrical vessel having diameter equal to its height is full of water which is poured into
two identical cylindrical vessels with diameter 42cm and height 21cm which are filled
completely. Find the diameter of cylindrical vessel?
Sol:
Given that diameter is equal to height of a cylinder
So 2h r
2Volume of cylinder r h
So volume 2 2r r
32 r
Volume of each vessel 2r h
Diameter 42cm
Height 21cm
Diameter 2d r
2 42r
21r
Radius 21cm
Volume of vessel 2
21 21 ……….(2)
Since volumes are equal
Equating (1) and (2)
232 21 21 2r ( 2 identical vessels)
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
2
321 21 2
2r
33 21r
21 42r d cm
Radius of cylindrical vessel 21cm
Diameter of cylindrical vessel 42 .cm
7. 50 circular plates each of diameter 14cm and thickness 0 5cm are placed one above other
to form a right circular cylinder. Find its total surface area?
Sol:
Given that 50 circular plates each with diameter 14cm
Radius of circular plates 7r cm
Thickness of plates 0 5
Since these plates are placed one above other so total thickness of plates 0 5 50
25 .cm
22 2Total surface area of acylinder rh r
22 2rh r
2 r h r
22
2 7 25 77
2. . 1408T S A cm
Total surface area of circular plates is 21408cm
8. 25 circular plates each of radius 10 5cm and thickness 1 6cm are placed one above the
other to form a solid circular cylinder. Find the curved surface area and volume of cylinder
so formed?
Sol:
Given that 25 circular plates each with radius (r) = 10.5cm
Thickness 1 6cm
Since plates are placed one above other so its height becomes 1 6 25 40cm
2Volume of cylinder r h
2
10 5 40
313860cm
2Curved surface area of a cylinder rh
2 10 5 40
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
222 10 5 40
7
22640cm
Volume of cylinder 313860cm
Curved surface area of a cylinder 22640cm
9. A path 2m wide surrounds a circular pond of diameter 4cm. how many cubic meters of
gravel are required to grave the path to a depth of 20cm
Sol:
Diameter of circular pond 40m
Radius of pond(r) 20 .m
Thickness 2m
Depth 20 0 2cm m
Since it is viewed as a hollow cylinder
Thickness t R r
2 R r
2 20R
22R m
2 2Volume of hollow cylinder R r h
2 222 20 h
2 222 20 0 2
84 0.2
352Volume of hollow cylinder m
352 77m of gravel is required to have path to a depth of 20cm.
10. A 16m deep well with diameter 3 5m is dug up and the earth from it is spread evenly to
form a platform 27 5m by 7m. Find height of platform?
Sol:
Let as assume well is a solid right circular cylinder
Radius of cylinder 3 5
1 752
r m
Height (or) depth of well 16 .m
2Volume of right circular cylinder r h
222
1 75 167
……..(1)
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Given that length of platform 27.5l m
Breath of platform 7b cm
Let height of platform be xm
tanVolume of rec gle lbh
27 5 7 192 5x x ………(2)
Since well is spread evenly to form platform
So equating (1) and (2)
1 2V V
222
1 75 16 192 57
x
0 8x m
( ) 80 .Height of platform h cm
11. A well of diameter 2m is dug14m deep. The earth taken out of it is spread evenly all around
it to form an embankment of height 40cm. Find width of the embankment?
Sol:
Let us assume well as a solid circular cylinder
Radius of circular cylinder2
12
m
Height (or) depth of well 14m
2Volume of solid circular cylindeer r h
2
1 14 …..(1)
Given that height of embankment (h) = 40cm
Let width of embankment be ‘x’ m
Volume of embankment 2r h
2
21 1 0 4x …..(2)
Since well is spread evenly to form embankment so their volumes will be same so equating
(1) and (2)
2
2 21 14 1 1 0 4x
5x m
5Width of embankment of x m
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
12. Find the volume of the largest right circular cone that can be cut out of a cube where edge
is 9cm_____?
Sol:
Given that side of cube =9cm
Given that largest cone is curved from cube
Diameter of base of cone = side of cube
2 9x
9
2r cm
Height of cone = side of cube
Height of cone (h) = 9cm
21arg
3Volume of l est cone r h
21 9
93 2
t
3912
3190 92cm
Volume of largest cone 3190 92v cm
13. A cylindrical bucket, 32 cm high and 18cm of radius of the base, is filled with sand. This
bucket is emptied on the ground and a conical heap of sand is formed. If the height of the
conical heap is 24 cm, find the radius and slant height of the heap.
Sol:
36cm, 43.27 cm
14. Rain water, which falls on a flat rectangular surface of length 6cm and breath 4m is
transferred into a cylindrical vessel of internal radius 10cm. What will be the height of
water in the cylindrical vessel if a rainfall of 1cm has fallen____?
Sol:
Given length of rectangular surface 6cm
Breath of rectangular surface 4cm
Height (h) 1cm
tanVolume of a flat rec gular surface lbh
6000 400 1
Volume 3240000cm _________(1)
Given radius of cylindrical vessel 20cm
Let height off cylindrical vessel be 1h
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Since rains are transferred to cylindrical vessel.
So equating (1) with (2)
2
1 1Volume of cylindrical vessel r h
2
1
2220
7h _________(2)
2
1
2224000 20
7h
1 190 9h cm
height of water in cylindrical vessel 190 9cms
15. A conical flask is full of water. The flask has base radius r and height h. the water is proved
into a cylindrical flask off base radius one. Find the height of water in the cylindrical flask?
Sol:
Given base radius of conical flask be r
Height of conical flask is h
21
3Volume of cone r h
So its volume 21
3r h ________(1)
Given base radius of cylindrical flask is ms.
Let height of flask be 1h
2
1Volume of cylinder r h
So its volume 2
1
22
7mr h _________(2)
Since water in conical flask is poured in cylindrical flask their volumes are same
(1) = (2)
22
1
1
3r h mr h
1 23
hh
m
Height of water in cylindrical flask 23
h
m
16. A rectangular tank 15m long and 11m broad is required to receive entire liquid contents
from a full cylindrical tank of internal diameter 21m and length 5m. Find least height of
tank that will serve purpose______?
Sol:
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Given length of rectangular tank = 15m
Breath of rectangular tank = 11m
Let height of rectangular tank be h
tan tanVolumeof rec gular k lbh
Volume 15 11 h ______(1)
Given radius of cylindrical tank 21
2r m
Length/height of tank = 5m
2tanVolume of cylindrical k r h
221
52
_______(2)
Since volumes are equal
Equating (1) and (2) 2
2115 11 5
2h
222 21
57 2
15 11h
10 5h m
Height of tank 10 5 .m
17. A hemisphere tool of internal radius 9cm is full of liquid. This liquid is to be filled into
cylindrical shaped small bottles each of diameter 3cm and height 4cm. how many bottles
are necessary to empty the bowl.
Sol:
Given that internal radius of hemisphere bowl 90m
34
3Volume of hemisphere r
32
93
______(1)
Given diameter of cylindrical bottle 3cm
Radius 3
2cm
Height 4cm
2Volume of cylindrical r h
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
23
42
______(2)
Volume of hemisphere bowl is equal to volume sum of n cylindrical bottles
(1) = (2)
2
32 39 4
3 2n
3
2
29
3
34
2
n
54n
No of bottles necessary to empty the bottle 54.
18. The diameters of the internal and external surfaces of a hollow spherical shell are 6 cm and
10 cm respectively. If it is melted and recast and recast into a solid cylinder of diameter 14
cm, find the height of the cylinder.
Sol:
Internal diameter of hollow spherical shell 6cm
Internal radius of hollow spherical shell 6
33
cm
External diameter of hollow spherical shell 10cm
External radius of hollow spherical shell10
52
cm
Diameter of cylinder 14 cm
Radius of cylinder 14
72
cm
Let height of cylinder xcm
According to the question
Volume of cylinder = Volume of spherical shell
2 3 34
7 5 33
x
4
49 125 273
x
449 98
3x
4 98 8
3 49 3x cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Height off cylinder 8
3cm
19. A hollow sphere of internal and external diameter 4cm and 8cm is melted into a cone of
base diameter 8cm. Calculate height of cone?
Sol:
Given internal diameter of hollow sphere (r) 4cm
External diameter 8R cm
2 24
3Volumeof hollow sphere R r
2 244
3B ____(1)
Given diameter of cone 8cm
Radius of cone 4cm
Let height of cone be h
21
3Volume of cone r h
21
43
h ____(2)
Since hollow sphere is melted into a cone so there volumes are equal
(1) = (2)
24 1
64 16 43 3
h
448
31
163
h
12h cm
Height of cone 12cm
20. A cylindrical tube of radius 12cm contains water to a depth of 20cm. A spherical ball is
dropped into the tube and the level of the water is raised by 6 75 .cm Find the radius of the
ball___?
Sol:
Given that radius of a cylindrical tube (r) 12cm
Level of water raised in tube 6 75h cm
2Volume of cylinder r h
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
2 312 6 75cm
2 322
12 6 257
cm ………(1)
Let ‘r’ be radius of a spherical ball
34
3Volume of sphere r ………(2)
To find radius of spherical balls
Equating (1) and (2)
2 34
12 6 753
r
2
312 6 75
4
3
r
3 729r 3 39r
9r cm
Radius of spherical ball 9r cm
21. 500 persons have to dip in a rectangular tank which is 80m long and 50m broad. What is
the rise in the level of water in the tank, if the average displacement of water by a person is 30 04m ____?
Sol:
Given that length of a rectangular tank (r) = 80m
Breath of a rectangular tank (b) 50m
Total displacement of water in rectangular tank
By 500 persons 3500 0 04m 320m ______(1)
Let depth of rectangular tank be h
tanVolume of rectangular k lbh
380 50 hm ______(2)
Equating (1) and (2)
20 80 50 h
20 4000h
20
4000h
0 005h m
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
0 5h cm
Rise in level of water in tank 0 05 .h cm
22. A cylindrical jar of radius 6cm contains oil. Iron sphere each of radius 1 5cm are immersed
in the oil. How many spheres are necessary to raise level of the oil by two centimetress?
Sol:
Given that radius of a cylindrical jar 6r cm
Depth/height of cylindrical jar 2h cm
Let no of balls be ‘n’
2Volume of a cylinder r h
2 3
1
226 2
7V cm ………(1)
Radius of sphere 1 5cm
34
3So volume of sphere r
3 3
2
4 221 5
3 7V cm ………(2)
Volume of cylindrical jar is equal to sum of volume of n spheres
Equating (1) and (2)
2 322 4 22
6 2 1 57 3 4
n
2
1
32
226 2
74 22
1 53 7
vn n
v
16n
No of spherical balls 16n
23. A hollow sphere of internal and external radii 2cm and 4cm is melted into a cone of basse
radius 4cm. find the height and slant height of the cone______?
Sol:
Given that internal radii of hollow sphere (r) = 2cm
External radii of hollow sphere 4R cm
2 24
3Volume of hollow sphere R r
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
2 2
1
44 2
3v …………(1)
Given that sphere is melted into a cone
Base radius of cone 4cm
Let slant height of cone be l
Let height of cone be h 2 2 2l r h 2 216l h …………(3)
21
3Volume of cone r h
2
2
14
3v h ………..(2)
1 2v v Equating (1) and (2)
22 24 1
4 2 43 3
h
416 4
31
163
h
14h cm
Substituting ‘h’ value in (2) 2 216l h 2 216 14l 2 16 196l
14 56l cm
Slant height of cone 14 56cm
24. The internal and external diameters of a hollow hemisphere vessel are 21cm and 25 2 .cm
The cost of painting 21cm of the surface is 10paise. Find total cost to paint the vessel all
over______?
Sol:
Given that internal diameter of hollow hemisphere 21
10 52
r cm cm
External diameter 25 2
12 62
R cm
Total surface area of hollow hemisphere
2 2 2 22 2R r R r
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
2 2 2 22 12 6 2 10 5 12 6 10 5
997 51 692 72 152 39 21843 38cm
Given that cost of painting 21cm of surface 10 ps
Total cost for painting 21843 38cm
1843 38 10ps
184 338 .Rs
Total cot to paint vessel all over 184 338 .Rs
25. A cylindrical tube of radius 12cm contains water to a depth of 20cm. A spherical ball of
radius 9cm is dropped into the tube and thus level of water is raised by hcm. What is the
value of h_____?
Sol:
Given that radius of cylindrical tube 1 12r cm
Let height of cylindrical tube (h)
2
1Volume of a cylinder r h
2
1 12v h ……(1)
Given spherical ball radius 2 9r cm
3
2
4
3Volume of sphere r
3
2
49
3v ……(2)
Equating (1) and (2)
1 2v v
2 34
12 93
h
3
2
49
3
12h
6 75h cm
Level of water raised in tube (h) 6 75cm
26. The difference between outer and inner curved surface areas of a hollow right circular
cylinder 14cm long is 288 .cm If the volume of metal used in making cylinder is 3176 .cm
find the outer and inner diameters of the cylinder____?
Sol:
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Given height of a hollow cylinder = 14cm
Let internal and external radii of hollow
Cylinder be ‘r’ and R
Given that difference between inner and outer
Curved surface 288cm
Curved surface area of cylinder (hollow)
22 R r h cm
88 2 R r h
88 2 14R r
1R r ……(1)
Volume of cylinder (hollow) 2 2 3R r h cm
Given volume of a cylinder 3176cm
2 2 176R r h
2 2 14 176R r
2 2 4R r
4R r R r
4R r ……(2)
1
4
2 5
R r
R r
R
52 5 2 5
2R R cm
Substituting ‘R’ value in (1)
1R r
2 5 1r
2 5 1 r
1 5r cm
Internal radii of hollow cylinder 1 5cm
External radii of hollow cylinder 2 5cm
27. Prove that the surface area of a sphere is equal to the curved surface area of the
circumference cylinder__?
Sol:
Let radius of a sphere be r
24Curved surface area of sphere r
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
2
1 4S r
Let radius of cylinder be ' 'r cm
Height of cylinder be '2 'r cm
2Curved surface area of cylinder rh
2
2 2 2 4S r r r
1S and 2S are equal. Hence proved
So curved surface area of sphere = surface area of cylinder
28. The diameter of a metallic sphere is equal to 9cm. it is melted and drawn into a long wire
of diameter 2mm having uniform cross-section. Find the length of the wire?
Sol:
Given diameter of a sphere (d) 9cm
Radius (r) 9
4 52
cm
34
3Volume of a sphere r
3 3
1
44 5 381 70
3V cm ……(1)
Since metallic sphere is melted and made into a cylindrical wire
2Volume of a cylinder r h
Given radius of cylindrical wire 2
2
mmr
1 0 1mm cm
2
2 0 1V h ……(2)
Equating (1) and (2)
1 2V V
2
381 703 0 1 h
12150h cm
Length of wire (h) 12150cm
29. An iron spherical ball has been melted and recast into smaller balls of equal size. If the
radius of each of the smaller balls is 1
4 of the radius of the original ball, how many such
balls are made? Compare the surface area, of all the smaller balls combined together with
that of the original ball.
Sol:
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Given that radius of each of smaller ball 1
4 Radius of original ball.
Let radius of smaller ball be .r
Radius of bigger ball be 4r
Volume of big spherical ball 34
3r 4r r
3
1
44
3V r ……(1)
34
3Volume of each small ball r
3
2
4
3V r ……(2)
Let no of balls be ' 'n
1
2
Vn
V
3
3
44
34
3
r
n
r
34 64n
64No of small balls
Curved surface area of sphere 24 r
Surface area of big ball 2
1 4 4S r …….(3)
Surface area of each small ball 2
1 4S r
Total surface area of 64 small balls
2
2 64 4S r …….(4)
By combining (3) and (4)
2 43
S
2 4S s
Total surface area of small balls is equal to 4 times surface area of big ball.
30. A tent of height 77dm is in the form a right circular cylinder of diameter 36m and height
44dm surmounted by a right circular cone. Find the cost of canvas at Rs.3.50 per 2m ?
Sol:
Given that height of a tent = 77dm
Height of cone 44dm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Height of a tent without cone 77 44 33dm
3 3m
Given diameter of cylinder (d) = 36m
Radius 36
182
r m
Let ‘l’ be slant height of cone 2 2 2l r h 2 2 218 3 3l 2 324 10 89l 2 334 89l
18 3l
Slant height of cone 18 3l
Curved surface area of cylinder 1 2S rh
22 18 4 4m …….(1)
Curved surface area of cone 2S rl
218 18 3m …….(2)
Total curved surface of tent 1 2S S
T.C.S.A 1 2S S
21532 46m
Given cost canvas per2 3 50m Rs
Total cost of canvas per 1532 46 3 50
1532 46 3 50
5363 61
Total cost of canvas 5363.61Rs
31. Metal spheres each of radius 2cm are packed into a rectangular box of internal dimension
16 8 8cm cm cm when 16 spheres are packed the box is filled with preservative liquid.
Find volume of this liquid?
Sol:
Given radius of metal spheres 2cm
34
3Volume of sphere v r
So volume of each metallic sphere 3 34
23
cm
Total volume of 16 spheres 3 3
1
416 2
3v cm …(1)
Volume of rectangular box lbh
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
3
2 16 8 8V cm …(2)
Subtracting (2) – (1) we get volume of liquid
2 1V V Volume off liquid
34
16 8 8 2 163
31024 536 16 488cm
Hence volume of liquid 3488cm
32. The largest sphere is to be curved out of a right circular of radius 7cm and height 14cm.
find volume of sphere?
Sol:
Given radius of cylinder 7r cm
Height of cylinder 14h cm
Largest sphere is curved out from cylinder
Thus diameter of sphere = diameter of cylinder
Diameter of sphere 2 7 14d cm
Volume of a sphere 34
3r
34
73
1372
3
31436 75cm
Volume of sphere 31436 75cm
33. A copper sphere of radius 3cm is melted and recast into a right circular cone of height 3cm.
find radius of base of cone?
Sol:
Given radius of sphere 3cm
Volume of a sphere 34
3r
3 343
3cm …..(1)
Given sphere is melted and recast into a right circular cone
Given height of circular cone 3 .cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Volume of right circular cone 2 1
3r h
2 23
3r cm
…..(1)
Equating 1 and 2 we get
234 1
3 33 3
r
3
2
43
3r
2 36r cm
6r cm
Radius of base of cone 6r cm
34. A vessel in the shape of cuboid contains some water. If these identical spheres are
immersed in the water, the level of water is increased by 2cm. if the area of base of cuboid
is 2160cm and its height 12cm, determine radius of any of spheres?
Sol:
Given that area of cuboid 2160cm
Level of water increased in vessel 2cm
Volume of a vessel 3160 2cm ……(1)
Volume of each sphere 3 34
3r cm
Total volume of 3 spheres 3 343
3r cm ……(2)
Equating (1) and (2) ( Volumes are equal 1 2V V )
34160 2 3
3r
3 160 2
43
3
r
3 320
4r
2 94r cm
Radius of sphere 2 94cm
35. A copper rod of diameter 1cm and length 8cm is drawn into a wire of length 18m of
uniform thickness. Find thickness of wire?
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Sol:
Given diameter of copper rod 1 1d cm
Radius 1
10 5
2r cm
Length of copper rod 1 8h cm
2
1 1Volume of cylinder r h
2 3
1 0 5 8V cm ……(1)
2
2 2 2V r h
Length of wire 2 18 1800h m cm
2 3
2 2 1800V r cm ..…..(2)
Equating (1) and (2)
1 2V V
2 2
20 5 8 1800r
2
2
2
0 5 8
1800r
2 0 033r cm
Radius thickness of wire 0 033 .cm
36. The diameters of internal and external surfaces of hollow spherical shell are 10cm and 6cm
respectively. If it is melted and recast into a solid cylinder of length of 2 2
3 cm, find the
diameter of the cylinder.
Sol:
Given diameter of internal surfaces of a hollow spherical shell 10cm
Radius 10
5 .2
r cm
External radii 6
32
R cm
2 24
3Volume of a spherica shell hollow R r
2 2 3
1
45 3
3V cm ………(1)
Given length of solid cylinder 8
3h
Let radius of solid cylinder be ' 'r
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
2Volume of a cylinder r h
2 3
2
8
3V r cm
………(2)
1 2V V
Equating (1) and (2)
24 825 9
3 3r
2
416
38
3
r
2 49r cm
7r cm
2 14d r cm
14Diameter of cylinder cm
37. A right angled triangle whose sides are 3 cm, 4 cm and 5 cm is revolved about the sides
containing the right angle in two days. Find the difference in volumes of the two cones so
formed. Also, find their curved surfaces.
Sol:
(i) Given that radius of cone 1 4r cm
Height of cone 1 3h cm
Slant height of cone 1 5l cm
Volume of cone 2
1 1 1
1
3V r h
2 31
4 3 163
cm
(ii) Given radius of second cone 2 3r cm
Height of cone 2 4h cm
Slant height of cone 2 5l cm
Volume of cone 2
2 2 2
1
3V r h
2 31
3 4 123
cm
Difference in volumes of two cones 1 2V V V
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
16 12V 34V cm
Curved surface area of first cone 1 1 1S rl
2
1 4 5 20S cm
Curved surface area of first cone 1 1 1S rl
2
1 4 5 20S cm
Curved surface area of second cone 2 2 2S r l
2
1 3 5 15S cm
2 2
1 220 15S cm S cm
38. How many coins 1 75cm in diameter and 2mm thick must be melted to form a cuboid
11 10 75 ___?cm cm cm
Sol:
Given that dimensions of a cuboid 11 10 75cm cm cm
So its volume 1 11 10 7V cm cm cm
311 10 7cm …….(1)
Given diameter (d) 1 75cm
Radius 1 75
0 8752 2
dr cm
Thickness 2 0 2h mm cm
2Volume of acylinder r h
2 3
2 0 875 0 2V cm …….(2)
1 2V V n
Since volume of a cuboid is equal to sum of n volume of ‘n’ coins
1
2
Vn
V
n no of coins
2
11 10 7
0 875 0 2n
1600n
No of coins 1600,n
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
39. A well with inner radius 4m is dug 14m deep earth taken out of it has been spread evenly
all around a width of 3m it to form an embankment. Find the height of the embankment?
Sol:
Given that inner radius of a well 4a m
Depth of a well 14h m
2Volume of a cylinder r h
2 3
1 4 14V cm ……….(1)
Given well is spread evenly to form an embankment
Width of an embankment 3m
Outer radii of a well (R) 4 3 7 .m
2 2 3Volumeof a hollow cylinder R r hm
2 2 3
2 7 4V hm ………..(2)
Equating (1) and (2)
1 2V V
2
4 14 49 16 h
24 14
33h
6 78h m
40. Water in a canal 1.5mwide and 6m deep is flowering with a speed of 10 / .km hr how much
area will it irrigate in 30 minutes if 8cm of standing water is desired?
Sol:
Given that water is flowering with a speed 10 /km hr
In 30 minutes length of flowering standing water 30
1060
km
5 5000 .km m
Volume of flowering water in 30 minutes 35000V width depth m
Given width of canal 1 5m
Depth of canal 6m 35000 1 5 6V m
345000V m
Irrigating area in 30 minutes if 8cm of standing water is desired 45000
0 08
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
245000562500
0 08m
230min 562500Irrigated area in utes m
41. A farmer runs a pipe of internal diameter 20 cm from the canal into a cylindrical tank in his
field which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of
3 km/h, in how much time will the tank be filled?
Sol: 9
8 𝑚
42. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly
all around it to a width of 4 m to form an embankment. Find the height of the embankment.
Sol:
Given diameter of well = 3m
Radius of well 3
42
m
Depth of well 14b m
With of embankment 4m
Radius of outer surface of embankment 3 11
42 2
m
Let height of embankment hm
2 2
1 2 1Volume of embankment V r r h
( it is viewed as a hollow cylinder) 2
2
3
1
11 3
2 2V h m
…..(1)
2
2 1Volume of earth dugout V r h
2
3
2
314
2V m
…..(2)
Given that volumes (1) and (2) are equal
So 1 2V V
2 2 211 3 3
142 2 2
h
121 9 914
4 4 4h
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
9
8h m
Height of embankment 9
.8
h m
43. The surface area of a solid metallic sphere is 616 cm2. It is melted and recast into a cone of
height 28 cm. Find the diameter of the base of the cone so formed (Use it = 22
7)
Sol:
Given height of cone 28h cm
Given surface area of Sphere 2616cm
We know surface area of sphere 24 r 24 616r
2 616 7
4 22r
2 49r
7r cm
Radius of sphere 7r cm
Let 1r be radius of cone
Given volume of cone Volume of sphere
21
3Volume of cone r h
2 3
1 1
128
3V r cm ………..(1)
3
2
4
3Volume of sphere V r
3 3
2
47
3V cm ………..(1)
(1) = (2) 1 2V V
2 3
1
1 428 7
3 3r
2
1 49r
1 7r cm
Radius of cone 1 7r cm
1 2 7 14Diameter of base of cone d cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
44. The difference between the outer and inner curved surface areas of a hollow right circular
cylinder 14cm long is 288cm . If the volume of metal used in making cylinder is 3176cm
find outer and inner diameters of the cylinder?
Sol:
Given height of a hollow cylinder 14cm
Let internal and external radii of hollow
Cylinder be ‘r’ an ‘R’
Given that difference between inner and outer curved surface 288cm
2Curved surface area of hollow cylinder R r h
88 2 0R h
88 2 14R r
1R r ……..(1)
2 2 3Volume of hollow cylinder R r h cm
Given volume of cylinder 3176cm
2 2 176R r h
2 2 14 176R r
2 2 4R r
4R r R r
4 4R ……..(2)
By using (1) and (2) equations and solving we get
1 ..(1)
4 ..(2)
2 5
R r
R r
R
52 5
2R cm
Substituting ‘R’ value in (1)
1
2 5 1
2 5 1
1 5
R r
r
r
r cm
External radii of hollow cylinder 2 5R cm
Internal radii of hollow cylinder 1 5r cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
45. The volume of a hemisphere is 312425 .
2cm Find its curved surface area?
Sol:
Given that volume of a hemisphere 312424
2cm
Volume of a hemisphere 32
3r
32 12425
3 2r
32 4841
3 2r
3 4851 3
2 2r
3 4851 3
4r
3r
10 50r cm
Radius of hemisphere 10 5cm
Curved surface area of hemisphere 22 r
2
2 10 5
692 72 2693xm
curved surface area off hemisphere 2693cm
46. A cylindrical bucket 32cm high and with radius of base 18cm is filled with sand. This
bucket is emptied out on the ground and a conical heap of sand is formed. If the height of
the conical heap of sand is formed. If the height of the conical heap is 24cm. find the radius
and slant height of the heap?
Sol:
Given that height of cylindrical bucket 32h cm
Radius 18r cm
Volume of cylinder 2r h
2 322
18 327
cm ………(1)
Given height of conical heap 24cm
Let radius of conical heap be 1r
Slant height of conical heap be 1l
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
2 2 2
1 1 1l r h
2 2 2
1 1 1r l h
22 2
1 1 24r l ………..(2)
Volume of cone 21
3r h
So its volume 2
1 1
1
3r h
2
1
1 2224
3 7r
2 3
1
228
7r cm ………….(3)
So equating (1) and (3)
(1) = (3)
2 2
1
22 2218 32 8
7 7r
2
2
1
18 32
8r
2
1 1296r
1 36r cm
Radius of conical heap is 36cm
Substituting 1r in (2)
22 2
1 1 24r l
2
11296 576l
2
11296 576 l
2
11872 l
1 43 26l cm
Slant height of conical heap 43 26cm
Exercise 16.2
47. A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of
cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone
is 16 m above the ground. Find the area of canvas required for the tent.
Sol:
Given diameter of cylinder 24m
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Radius 24
122
r m
Given height of cylindrical part 1 11h m
Height of cone part 2 5h m
Vertex of cone above ground 11 5 16m
Curved surface area of cone 1S rl
2212
7l
Let l be slant height of cone
2 2
2l r h
2 212 5 13l m
13l m
Curved surface area of cone 2225 12 13
7m ……….(1)
Curved surface area of cylinder 2 2S rh
2
2 2 12 11S m ………..(2)
To find area of canvas required for tent
1 2 1 2S S S
22
12 13 2 12 117
S
490 829 38S 21320S m
Total canvas required for tent 21320S m
48. A rocket is in the form of a circular cylinder closed at the lower end with a cone of the
same radius attached to the top. The cylinder is of radius 2 5m and height 21m and the
cone has a slant height 8m. Calculate total surface area and volume of the rocket?
Sol:
Given radius of cylinder 2 5a m
Height of cylinder 21h m
Slant height of cylinder 8l m
Curved surface area of cone 1S rl
2
1 2 5 8S cm ………(1)
Curbed surface area of a cone 22 rh r
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
2 2
2 2 2 5 21 2 5S cm ……….(2)
Total curved surface area = (1) + (2)
1 2S S S
2
2 5 8 2 2 5 21 2 5S
62 831 329 86 19 63S 2412 3S m
Total curved surface area 2412 3m
Volume of a cone 21
3r h
2 3
1
12 5
3V h cm ………...(3)
Let ‘h’ be height of cone 2 2 2l r h
2 2 2l r h
2 2h l r
2 28 25h
23 685h m
Subtracting ‘h’ value in (3)
Volume of a cone 2 2
1
12 5 23 685
3V cm ………(4)
Volume of a cylinder 2
2V r h
2 32 5 21m ………..(5)
Total volume = (4) + (5)
1 2V V V
2 21
2 5 23 685 2 5 13
V
2461 84V m
Total volume 2461 84V m
49. A tent of height 77 dm is in the form of a right circular cylinder of diameter 36 m and
height 44 dm surmounted by a right circular cone. Find the cost of the canvas at Rs. 350
per m2 (Use it = 22
7).
Sol:
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Given that height of a tent 77dm
Height of a surmounted cone 44dm
Height of cylinder part 77 44
33 3.3dm m
Given diameter of cylinder (d) 26m
Radius 36
18 .2
r m
Let ‘l’ be slant height of cone 2 2 2l r h 2 2 218 3 3l 2 824 10 89l
18 3l
Slant height of cone 1 18 3
Curved surface area of cylinder 1 2S rh
22 18 4 4m ………..(1)
Curved surface area of cone 2S rh
218 18 3m ………..(2)
Total curved surface of tent 1 2S S
1 2S S S
21532 46S m
Total curved surface area 212 1532 46S m
50. A toy is in the form of a cone surmounted on a hemisphere. The diameter of the base and
the height of cone are 6cm and 4cm. determine surface area of toy?
Sol:
Given height of cone 4h cm
Diameter of cone 6d cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Radius (r) 6
32
cm
Let ‘l’ be slant height of cone
2 2l r h
2 23 4 5cm
5l cm
Slant height of cone 5 .l cm
Curved surface area of cone 1S rl
2
1 3 5 47 1S cm
Curved surface area of hemisphere 2
2 2S r
2 2
2 2 3 56 52S cm
Total surface area 1 26s S
47 1 56 52 2103 62cm
Curved surface area of toy 2103 62cm
51. A solid is in the form of a right circular cylinder, with a hemisphere at one end and a cone
at the other end. The radius of the common base is 3.5 cm and the heights of the cylindrical
and conical portions are 10 cm. and 6 cm, respectively. Find the total surface area of the
solid. (Use n = 22
7)
Sol:
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Given radius of common base 3 5cm
Height of cylindrical part 10h cm
Height of conical part 6h cm
Let ' 'l be slant height of cone
2 2l r h
2 23 5 6l
48 25l cm
Curved surface area of cone 1S rl
3 5 48 25
276 408cm
Curved surface area of cylinder 2 2S rh
2 3 5 10
2220cm
Curved surface area of hemisphere 1 2 3S S S S
76 408 220 77 2373 408cm
Total surface area of solid 2373 408S cm
Cost of canvas per 2 3 50m Rs
Cost of canvas for 21532 46 1532 46 3 50m
5363 61Rs
Cost of canvas required for tent 5363 61Rs pr
52. A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone
on the other. The radius and height of the cylindrical part are 5 cm and 13 cm respectively.
The radii of the hemispherical and conical parts are the same as that of the cylindrical part.
Find the surface area of the toy if the total height of the toy is 30 cm.
Sol:
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
1 2 2 13S
2
1 408 2S cm
Curved surface area of cone 2S rl
Let l be slant height of cone
2 2l r h
30 13 5 12h cn
2 212 5 13l cm
13l cm
Curved surface area of cone 2 5 13S
2204 1cm
Curved surface area of hemisphere 2
3 2S r
2
2 5
22 25 50 157cm
2
3 157S cm
Total curved surface area 1 2 3S S S S
408 2 204 1 157S 2769 3S cm
Surface area of toy 2769.3S cm
53. A cylindrical tube of radius 5cm and length 9 8cm is full of water. A solid in form of a
right circular cone mounted on a hemisphere is immersed in tube. If radius of hemisphere is
immersed in tube if the radius of hemisphere is 3 5cm and height of the cone outside
hemisphere is 5cm. find volume of water left in the tube?
Sol:
Given radius of cylindrical tube 5 .r cm
Height of cylindrical tube 9 8h cm
Volume of cylinder 2r h
2 3
1 5 9 8 770V cm
Given radius of hemisphere 3 5r cm
Height of cone 5h cm
Volume of hemisphere 32
3r
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
3 32
3 5 89 793
cm
Volume of cone 21
3r h
2 33 5 5 64 14
3cm
Volume of cone + volume of hemisphere 3
2 39 79 64 14 154V cm
54. A circular tent has cylindrical shape surmounted by a conical roof. The radius of cylindrical
base is 20 .m The height of cylindrical and conical portions are 4 2m and 2 1 .m Find the
volume of the tent?
Sol:
Given radius of cylindrical base 20m
Height of cylindrical part 4 2 .h m
Volume of cylindrical 2
1r h
2 3
1 20 4 2 5280V m
Volume of cone 2
2
1
3r h
Height of conical part 2 2 1h m
2 3
2 20 2 1 8803
V m
Volume of tent 1 2v V V
5280 880V 36160V m
Volume of tent 1 2v V V
5280 880V 36160V m
Volume of tent 36160v m
55. A petrol tank is a cylinder of base diameter 21cm and length 18cm fitted with conical ends
each of axis 9cm. determine capacity of the tank?
Sol:
Given base diameter of cylinder 21cm
Radius 21
11 52
r cm
Height of cylindrical part 18h cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Height of conical part 2 9h cn
Volume of cylinder 2
1r h
2 3
1 11 5 18 7474 77V cm
Volume of cone 2
2
1
3r h ( 2 conical end)
2
2
111 5 9 2
3V
3
2
11190 25 2492 25
3V cm
Volume of tank = volume of cylinder + volume of cone
1 2V V V
7474 77 2492 85V 39966 36V cm
Volume of water left in tube = Volume of cylinder – Volume of hemisphere and cone
1 2V V V
770 154 3616cm
Volume of water left in tube 3616cm
56. A conical hole is drilled in a circular cylinder of height 12cm and base radius 5cm. The
height and base radius of the cone are also the same. Find the whole surface and volume of
the remaining cylinder?
Sol:
Given base radius of cylinder 5r cm
Height of cylinder 12h cm
Let ‘l’ be slant height of cone
2 2l r h
2 25 12
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
13l cm
Height and base radius of cone and cylinder are same
Total surface area of remaining part 22s rh r rl
2
2 5 12 5 5 13
T.S.A 2210 cm
Volume of remaining part = Volume of cylinder – Volume of cone
2 21
3V r h r h
2 21
5 12 5 123
V
3200V cm
Volume of remaining part 3200v cm
57. A tent is in form of a cylinder of diameter 20m and height 2 5m surmounted by a cone of
equal base and height 7 5 .m Find capacity of tent and cost of canvas at Rs 100per square
meter?
Sol:
Given radius of cylinder 20
102
r m
Height of a cylinder 1 2 5h m
Height of cone 2 7 5h m
Let ‘l’ be slant height of cone
2 2
2l r h
2 210 7 5l
12 5l m
Volume of cylinder 2
1V r h
2
1 10 2 5V ……..(1)
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Volume of cone 2
2 2
1
3V r h
2 31
10 7 53
m ……..(2)
Total capacity of tent = (1) + (2)
1 2V V V
2 21
10 2 5 10 7 53
V
250 250V 3500V cm
Total capacity of tent 2500 cm
58. A boiler is in the form of a cylinder 2m long with hemispherical ends each of 2m diameter.
Find the volume of the boiler?
Sol:
Given height of cylinder 2h m
Diameter of hemisphere 2d m
Radius 1r m
Volume of a cylinder 2r h
2 3
1 1 2V cm ……..(1)
Volume of hemisphere 32
3r
Since at ends of cylinder hemisphere are attached
Volumes of 2 hemispheres
2 22
2 13
cm ………(2)
Volumes of boiler = (1) + (2)
1 2V V V
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
2 22
2 1 1 23
V
3220
21V m
Volumes of boiler 3220
21m
59. A vessel is a hollow cylinder fitted with a hemispherical bottom of the same base. The
depth of cylinder is 14
3m and internal surface area of the solid?
Sol:
Given radius of hemisphere 3 5
1 752
r m
Height of cylinder 14
3h m
Volume of cylinder 2r h
2 314
1 753
cm
……..(1)
Volume of hemisphere 32
3r
3 32
1 753
cm ……..(2)
Volume of vessel = (1) + (2)
1 2V V V
2 32
3V r h r
2 214 2
1 75 1 753 3
V
356V m
Volumes of vessel 356v m
Internal surface area of solid 22 2s rh r
S = Surface area of cylinder + surface are of hemisphere
S = 214
2 1 75 2 1 753
270 51S m
Internal surface area of solid 270 51s m
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
60. A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid
is 104cm and radius of each of hemispherical ends is 7cm. find the cost of polishing its
surface at the rate of 210Rs per dm ?
Sol:
Given radius of hemispherical ends 7cm
Height of body 2 104 .h r cm
Curved surface area of cylinder 2 rh
2 7 h ………(1)
2 104h x
104 2h r
90h cm
Substitute ‘h’ value in (1)
Curved surface area of cylinder 2 7 90
23948 40cm ………(2)
Curved surface area of 2 hemisphere 22 2 r
22 2 7
3615 75cm ………(3)
Total curved surface area = (2) + (3) 2 23958 40 615 75 4574 15 45 74cm dm
Cost of polishing for 21 10dm Rs
Cost of polishing for 245 74 45 74 10dm
457 4Rs
61. A cylindrical vessel of diameter 14cm and height 42cm is fixed symmetrically inside a
similar vessel of diameter 16cm and height 42 .cm The total space between two vessels is
filled with cork dust for heat insulation purpose. How many cubic cms of cork dust will be
required?
Sol:
Given height of cylindrical vessel 42h cm
Inner radius of a vessel 1
147
2r cm cm
Outer radius of a vessel 2
168
2r cm
Volume of a cylinder 2 2
2 1r r h
2 28 7 42
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
64 49 42
15 42
630 31980cm
Volume of a vessel 21980cm
62. A cylindrical road solar made of iron is 1m long its internal diameter is 54cm and thickness
of the iron sheet used in making roller is 9cm. Find the mass of roller if 31cm of iron has
7 8gm mas?
Sol:
Given internal radius of cylindrical road
Roller 1
5427
2r cm
Given thickness of road roller 1
9cmb
Let order radii of cylindrical road roller be R
t R r
9 27R
9 27 36R cm
36R cm
Given height of cylindrical road roller 1h m
100 .h cm
Volume of iron 2 2h R r
2 236 27 100
31780 38cm
Volume of iron 31780 38cm
Mass of 31cm of iron 7 8gm
Mass of 31780 38cm of iron 1780 38 7 8
1388696 4gm
1388 7kg
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Mass of roller 1388 7m kg
63. A vessel in from of a hollow hemisphere mounted by a hollow cylinder. The diameter of
hemisphere is 14cm and total height of vessel is 13cm. find the inner surface area of
vessel?
Sol:
Given radius of hemisphere and cylinder (r)
147
2cm
Given total height of vessel 13cm
13h r cm
Inner surface area of vessel 2 r h r
2 7 13
182 2572cm
Inner surface area of vessel 2572cm
64. A toy is in the form of a cone of radius 3 5cm mounted on a hemisphere of same radius.
The total height of toy is 15 5 .cm Find the total surface area of toy?
Sol:
Given radius of cone 3 5r cm
Total height of toy 15 5h cm
Length of cone 15 5 3 5l
12cm
Length of cone 12l cm
Curved surface area of cone rl
1 3 5 12S
2
1 131 94S cm ……(1)
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Curved surface area of hemisphere 22 r
2
2 2 3 5S
2
2 76 96S cm ……(2)
Total surface of toy = (1) + (2)
1 2S S S
181 94 76 96S
208 90S 2209S cm
Total surface area of toy 2209cm
65. The difference between outside and inside surface areas of cylindrical metallic pipe 14cm
long is 244m . If pipe is made of 399cm of metal. Find outer and inner radii of pipe?
Sol:
Let inner radius of pipe be 1r
Radius of outer cylinder be 2r
Length of cylinder 14 .h cm
Surface area of hollow cylinder 2 12 h r r
Given surface area of cylinder 244m
66. A radius circular cylinder bring having diameter 12cm and height 15cm is full ice-cream.
The ice-cream is to be filled in cones of height 12cm and diameter 6cm having a
hemisphere shape on top find the number of such cones which can be filled with ice-
cream?
Sol:
Given radius of cylinder 1
126
2r cm
Given radius of hemisphere 2
63 .
2r cm
Given height of cylinder 15 ..h cm
Height of cones 12 .l cm
Volume of cylinder 2
1r h
2 36 15 cm ……(1)
Volume of each cone = volume of cone + volume of hemisphere
2 3
2 2
1 2
3 3r l r
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
2 3 31 2
3 12 33 3
cm ……..(2)
Let number of cones be ‘n’
n(Volume of each cone) = volume of cylinder
2 3 21 2
3 12 3 6 153 3
n
2
2 3
6 15
1 23 12 3
3 3
n
54010
5n
2 12 44h r r
2 12 14 44r r
2 128 44r r
2 1
44
28r r
2 1
1
2r r ………..(1)
Given volume of a hollow cylinder 399cm
Volume of a hollow cylinder 2 2
2 1h r r
2 2
2 1 99h r r
2 2
2 114 99r r
1 2 2 114 99r r r r
1 214 1 99r r
1 214 99r r
1 2
9
2r r ………..(2)
Equating (1) and (2) equations we get
1 2
1 2
2
9
2
1
2
2 5
r r
r r
r
2
5.
2r cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Substituting 2r value in (1)
1 2r cm
Inner radius of pipe 2a cm
Radius of outer cylinder 2
5.
2r cm
67. A solid iron pole having cylindrical portion 110cm high and of base diameter 12cm is
surmounted by a cone 9cm high. Find the mass of the pole given that the mass of 31cm of
iron is 8 ?gm
Sol:
Given radius of cylindrical part 12
62
r cm
Height of cylinder 110h cm
Length of cone 9l cm
Volume of cylinder 2r h
2 3
1 0 110V cm ………(1)
Volume of cone 21
3r l
2 3
2
16 9 108
3V cm ………(2)
Volume of pole = (1) + (2)
1 2V V V
2
6 110 108V
312785 14V cm
Given mass of 31cm of iron 8gm
Mass of 312785 14cm of iron 12785 14 8
102281 12
102 2kg
Mass of pole for 312785 14cm of iron is 102 2kg
68. A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of
the cone is 2 cm and the diameter of the base is 4 cm. If a right circular cylinder
circumscribes the toy, find how much more space it will cover.
Sol:
Given radius of cone, cylinder and hemisphere 4
22
r cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Height of cone 2l cm
Height of cylinder 4h cm
Volume of cylinder 22 32 4r h cm ………(1)
Volume of cone 21
3r l
21
2 23
34 23
cm
……….(2)
Volume of hemisphere 32
3r
32
23
328
3cm ………..(3)
So remaining volume of cylinder when toy is inserted to it 2 2 31 2
3 3r h r l r
= (1) – ((2) + (3))
2 2
2 4 8 83 3
3216 4 8 16 8 8
3cm
So remaining volume of cylinder when toy is inserted to it 38 cm
69. A solid consisting of a right circular cone of height 120cm and radius 60cm is placed
upright in right circular cylinder full of water such that it touches bottoms. Find the volume
of water left in the cylinder. If radius of cylinder is 60cm and its height is 180 ?cm
Sol:
Given radius of circular cone 60a cm
Height of circular cone 120 .b cm
Volume of a cone 21
3r l
2 31
60 1203
cm …………(1)
Volume of hemisphere 32
3r
Given radius of hemisphere 60cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
2 32
603
cm ………….(2)
Given radius of cylinder 60cm
Height of cylinder 180 .h cm
Volume of cylinder 2r h
2 360 180cm ……………(3)
Volume of water left in cylinder = (3) – ((1) + (2))
3 3 21 2
60 120 60 60 1803 3
3 3113 1 1 131cm m
Volume of water left in cylinder 31 131m
70. A cylindrical vessel with internal diameter 10cm and height 10 5cm is full of water. A
solid cone of base diameter 7cm and height 6cm is completely immersed in water. Find
value of water (i) displaced out of the cylinder (ii) left in the cylinder?
Sol:
Given internal radius 1
105
2r cm
Height of cylindrical vessel 10 5h cm
Outer radius of cylindrical vessel 2
73 5
2l cm
Length of cone 6 .l cm
(i) Volume of water displaced = volume of cone
Volume of cone 2
2
1
3r l
2 313 5 6 76 9
3cm
377cm
Volume of water displaced 377cm
Volume of cylinder 22
1 5 10 5r h
824 6 2825cm
(ii) Volume of water left in cylinder = volume of
Cylinder – volume of cone 3825 77 748cm
Volume of water left in cylinder 3748cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
71. A hemispherical depression is cut from one face of a cubical wooden block of edge 21cm
such that the diameter of hemisphere is equal to the edge of cube determine the volume and
total surface area of the remaining block?
Sol:
Given edge of wooden block 21a cm
Given diameter of hemisphere = edge of cube
Radius 21
10 52
cm
Volume of remaining block = volume of box – volume of hemisphere
3 32
3a r
3 32
2 10 53
36835 5cm
Surface area of box 26a ……..(1)
Curved surface area of hemisphere 22 r ..……(2)
Area of base of hemisphere 2r ……..(3)
So remaining surface area of box = (1) – (2) + (3) 2 2 26 2a r r
2 2
6 21 10 5 2 10 5
22992 5cm
Remaining surface area of box 22992 5cm
Volume of remaining block 36835 5cm
72. A tag is in the form of a hemisphere surmounted by a right circular cone of same base
radius as that of the hemisphere. If the radius of the base of cone is 21cm and its volume is
2
3of volume of hemisphere calculate height of cone and surface area of toy?
Sol:
Given radius of cone = radius of hemisphere
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Radius 21r cm
Given that volume of cone 2
3 Volume of hemisphere
Volume of cone 21
3r h
Volume of hemisphere 32
3r
So 2 31 2 2
3 3 3r h r
2 31 2 2
21 213 3 3
h
4 21 3
4 21h
421 28
3h cm
Height of cone 28h cm
Curved surface area of cone rl
2
1 21 28S cm ………….(1)
Curved surface area off hemisphere 22 r
2 2
2 2 21S cm …………..(2)
Total surface area 1 2 1 2s S S
22S rl r
25082S cm
Curved surface area of toy 25082cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
73. A solid is in the shape of a cone surmounted on hemisphere the radius of each of them is
being 3 5cm and total height of solid is 9 5cm . Find volume of the solid?
Sol:
Given radius of hemisphere and cone 3 5cm
Given total height of solid 9 5h cm
Length of cone 9 5 3 5 6l cm
Volume of a cone 21
3r l
2 3
1
13 5 6
3V cm …………(1)
Volume of hemisphere 32
3r
3 3
2
23 5
3V cm ………………..(2)
Volume of solid = (1) + (2)
1 2V V V
2 31 2
3 5 6 3 53 3
V
376 96 89 79 166 75V cm
Volume of solid 3166 75v cm
Exercise 16.3
1. A bucket has top and bottom diameters of 40 cm and 20 cm respectively. Find the volume of
the bucket if its depth is 12 cm. Also, find the cost of tin sheet used for making the bucket at
the rate of Rs 1.20 per dm2 . (Use 𝜋 = 3.14)
Sol:
Given diameter to top of bucket = 40cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Radius 1
4020
2r cm
Depth of a bucket 12h cm
Volume of a bucket 2 2
1 2 1 2
1
3r r r r h
12
2 2320 10 20 10
1
38800 .cm
Let ‘l’ be slant height of bucket
2
1 2 2l r r h
2 220 10 12l
2 61 15 620l cm
Total surface area of bucket 2
1 2 2r r l r
2
20 10 15 620 10
21320 61 2200
7cm
2 21320 61 220017 87
7 100dm dm
Given that cost of tin sheet used for making bucket per 2 1.20dm Rs
So total cost for 217 87 1 20 17 87dm
21 40 .Rs
Cost of tin sheet for 217 87 2140dm Rs ps
2. A frustum of a right circular cone has a diameter of base 20cm, of top 12cm and height
3cm. find the area of its whole surface and volume?
Sol:
Given base diameter of cone 1 20d cm
Radius 1
2010
2r cm
Top diameter of cone 2 12d cm
Radius 2
126
2r cm
Height of cone 3h cm
Volume of frustum right circular cone
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
2 2
1 2 1 2
1
3r r r r h
2 2110 6 10 6 3
3
3616cm
Let ‘l’ be slant height of cone
2
1 2 2l r r h
2 210 6 3l
16 9 25 5l cm cm
Slant height of cone 5l cm
Total surface area of cone 2 2
1 2 1 2r r l r r
2 2
10 6 5 10 6
80 100 36
2216 678 85cm
Total surface area of cone 2678 85cm
3. The slant height of the frustum of a cone is 4cm and perimeters of it circular ends are 18cm
and 6cm. find curved surface of the frustum?
Sol:
Given slant height of cone 4r cm
Let radii of top and bottom circles be 1r and 2r
Given perimeters of its ends as 18cm and 6cm
12 18r cm
1 9r cm ……(1)
22 6r cm
2 3r cm ……(2)
Curved surface area of frustum cone 1 2r r l
1 2r r l
1 2r r l
9 3 4
212 4 48cm
Curved surface area of frustum cone 248cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
4. The perimeters of the ends of a frustum of a right circular cone are 44 cm and 33 cm. If the
height of the frustum be 16 cm, find its volume, the slant surface and the total surface.
Sol:
Given perimeters of ends of frustum right circular cone are 44cm an 33cm
Height of frustum cone 16cm
Perimeter 2 r
12 44r
1 7r cm
22 33r
2
215 25
4r cm
Let slant height of frustum right circular cone be l
2 2
1 2l r r h
2 27 5 25 16l cm
16 1l cm
Slant height of frustum cone 16 1cm
Curved surface area of frustum cone 1 2r r l
7 5 25 16 1
C.S.A of cone 2619 65cm
Volume of a cone 2 2
1 2 1 2
1
3r r rr h
2217 5 25 7 5 25 16
3
31898 56cm
Volume of a cone 31898 56 cm
Total surface area of frustum cone 2 2
1 2 1 2r r l r r
2 27 5 25 16 1 7 5 25
2860 27cm
Total surface area of frustum cone 2860 27cm
5. If the radii of circular ends of a conical bucket which is 45cm high be 28cm and 7cm. find
the capacity of the bucket?
Sol:
Given height of conical bucket 45cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Give radii of 2 circular ends of a conical bucket is 28cm and 7cm
1 28r cm
2 7r cm
Volume of a conical bucket 2 2
1 2 1 2
1
3r r r r h
2 2128 7 28 7 45
3
1
1029 453
15435 348510V cm
Volume of a conical bucket 348510cm
6. The height of a cone is 20cm. A small cone is cut off from the top by a plane parallel to the
base. If its volumes be 1
25of the volume of the original cone, determine at what height
above base the section is made
Sol:
V AB be a cone of height 1 1 20h VO cm
Fronts triangles 1Vo A and 1VoA
1 1 1
1 1
20VO O A O A
VO OA VO OA
Volumes of cone 1
1
125VAO times volumes of cone VAB
We have 2 2
1 1 1
1 1 120
3 125 3OA VO O A
2
1
1
4
25
OAVO
O A
24
20 25
VOVO
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
3 4 400
25VO
3 64VO
4VO
Height at which section is made 20 4 16 .cm
7. If the radii of circular ends of a bucket 24cm high are 5cm and 15cm. find surface area of
bucket?
Sol:
Given height of a bucket 24R cm
Radius of circular ends of bucket 5cmand 15cm
1 25 ; 15r cm r cm
Let ' 'l be slant height of bucket
2 2
1 2l r r h
2 215 5 24l
100 576 676l
26l cm
Curved surface area of bucket 2
1 2 2r r l r
2
5 15 26 15
2
20 26 15
520 225
2745 cm
Curved surface area of bucket 2745 cm
8. The radii of circular bases of a frustum of a right circular cone are 12cm and 3cm and
height is 12cm. find the total surface area volume of frustum?
Sol:
Let slant height of frustum cone be ‘l’
Given height of frustum cone 12cm
Radii of a frustum cone are 12cm and 23cm
1 212 3r cm r cm
2 2
1 2l r r h
2 212 3 12l
81 144 15l cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
15l cm
Total surface area of cone 2 2
1 2 1 2r r l r r
2 2
12 3 15 12 3
T.S.A 2378 cm
Volume of cone 2 2
1 1 2 2
1
3r rr r h
2 2112 3 12 3 12
3
3756 cm
Volume of frustum cone 3756 cm
9. A tent consists of a frustum of a cone copped by a cone. If radii of ends of frustum be 13m
and 7m the height of frustum be 8m and slant height of thee conical cap bee 12m. find
canvas required for tent?
Sol:
Given height of frustum 8h m
Radii of frustum cone are 13 7m and m
1 213 7r m r cm
Let ‘l’ be slant height of frustum cone
2 2
1 2l r r h
2 213 7 8 36 64l
10l m
Curved surface area of friction 1 1 2S r r l
13 7 10
2200 m
C.S.A of frustum 2
1 200S m
Given slant height of conical cap = 12m
Base radius of upper cap cone 7m
Curved surface area of upper cap cone 2S rl
27 12 264m
Total canvas required for tent 1 2S S S
2200 264 892 57S m
Total canvas 2892 57m
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
10. A reservoir in form of frustum of a right circular contains 744 10 liters off water which
fills it completely. The radii of bottom and top of reservoir are 50m and 100m. find depth
of water and lateral surface area of reservoir?
Sol:
Let depth of frustum cone be h
Volume of first cone 2 2
1 2 1 2
1
3V r r rr h
1 250 100r m r m
2 21 2250 100 50 100
3 7V h
1 22
2500 1000 50003 7
V h …(1)
Volumes of reservoir 744 10 liters ….(2)
Equating (1) and (2)
212500 44 10
3h
24h
Let ‘l’ be slant height of cone
2
1 2 2l r r h
2 250 100 24l
55 461l m
Lateral surface area of reservoir
1 2S r r l
50 100 55 461
21500 55 461 26145 225m
Lateral surface area of reservoir 226145 225m
Volume of frustum cone 2 2
1 2 1 2
1
3r r r r h
2 2130 18 30 18 9
3
35292 cm
Volume 35292 cm
Total surface area of frustum cone =
2 2
1 2 1 2r r l r r
2 2
30 18 15 30 18
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
2 248 15 30 18
720 900 324
21944 cm
Total surface area 21944 cm
11. A metallic right circular cone 20cm high and whose vertical angle is 90 is cut into two
parts at the middle point of its axis by a plane parallel to base. If frustum so obtained bee
drawn into a wire of diameter 1
.16
cm
find length of the wire?
Sol:
Let ABC be cone. Height of metallic cone 20AO cm
Cone is cut into two parts at the middle point of its axis
Hence height of frustum cone 10AD cm
Since angle A is right angled. So each angles B and C 45
Angles E and F 45
Let radii of top and bottom circles of frustum cone bee 1r and 2r cm
From cot 45le DEADE
AD
1 110
r
1 10 .r cm
From le AOB
cot 45OB
OA
2 120
r
2 20r cm
12. A bucket is in the form of a frustum of a cone with a capacity of 12308.8 cm3 of water.
The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the
height of the bucket and the area of the metal sheet used in its making. (Use 𝜋 = 3.14).
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Sol:
Given radii of top circular ends 1 20r cm
Radii of bottom circular end of bucket 2 12r cm
Let height of bucket be ‘h’
Volume of frustum cone 2 2
1 2 1 2
1
3r r r r h
2 2120 12 20 12
3h
3784
3hcm ……….(1)
Given capacity/volume of bucket 3123308 8cm ……….(2)
Equating (1) and (2)
78412308 8
3h
12308 8 3
784h
15h cm
Height of bucket 15h cm
Let ‘l’ be slant height of bucket
22 2
1 2l r r h
2 2
1 2l r r h
2 220 2 15 64 225l
17l cm
Length of bucket/ slant height of
Bucket 17l cm
Curved surface area of bucket 2
1 2 2r r l r
2
20 12 17 12
2
32 17 12
29248 144 2160 32cm
Curved surface area 22160 32cm
13. A bucket made of aluminum sheet is of height 20cm and its upper and lower ends are of
radius 25cm an 10cm, find cost of making bucket if the aluminum sheet costs Rs 70 per 2100cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Sol:
Given height of bucket 20h cm
Upper radius of bucket 1 25r cm
Lower radius of bucket 2 10r cm
Let ‘l’ be slant height of bucket
2 2
1 2l r r h
2 225 10 20 225 400l
25l m
Slant height of bucket (l) 25cm
Curved surface area of bucket 2
1 2 2r r l r
2
25 10 25 10
35 25 100 975
C.S.A 23061 5cm
Curved surface area 23061 5cm
Cost of making bucket per 2100 70cm Rs
Cost of making bucket per 2 3061 53061 5 70
100cm
2143 05Rs
Total cost for 23061 5 2143 05cm Rs per
14. Radii of circular ends of a solid frustum off a cone re 33cm and 27cm and its slant height
are 10cm. find its total surface area?
Sol:
Given slant height of frustum cone 10cm
Radii of circular ends of frustum cone are 33and 27cm
1 233 ; 27 .r cm r cm
Total surface area of a solid frustum of cone
2 2
1 2 1 2r r l r r
2 2
33 27 10 33 27
2 2
60 10 33 27
600 1089 729
22418 cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
27599 42cm
Total surface area of frustum cone 27599 42cm
15. A bucket made up of a metal sheet is in form of a frustum of cone of height 16cm with
diameters of its lower and upper ends as 16cm and 40cm. find thee volume of bucket. Also
find cost of bucket if the cost of metal sheet used is Rs 20 per 100 2cm
Sol:
Given height off frustum cone 16cm
Diameter of lower end of bucket 1 16d cm
Lower and radius 1
168
2r cm
Upper and radius 2
4020
2r cm
Let ‘l’ be slant height of frustum of cone
2 2
1 2l r r h
2 220 8 16l
144 256l
20l cm
Slant height of frustum cone 20 .l cm
Volume of frustum cone 2 2
1 2 1 2
1
3r r r r h
2 218 20 8 20 16
3
1
99843
Volume 310449 92cm
Curved surface area of frustum cone
2
1 2 2r r l r
2
20 8 20 8
2560 64 624 cm
Cost of metal sheet per 2100 20cm Rs
Cost of metal sheet for 2 624624 20
100cm
391 9Rs
Total cost of bucket 391 9Rs
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
16. A solid is in the shape of a frustum of a cone. The diameter of two circular ends are 60cm
and 36cm and height is 9cm. find area of its whole surface and volume?
Sol:
Given height of a frustum cone = 9cm
Lower end radius 1
6030
2r cm cm
Upper end radius 2
3618
2r cm cm
Let slant height of frustum cone be l
2 2
1 2l r r h
2 28 30 9l
144 81l
15l cm
Volume of frustum cone 2 2
1 2 1 2
1
3r r r r h
2 2130 18 30 18 9
3
35292 cm
Volume 35292 cm
Total surface area of frustum cone =
2 2
1 2 1 2r r l r r
2 2
30 18 15 30 18
2 248 15 30 18
720 900 324
21944 cm
Total surface area 21944 cm
17. A milk container is made of metal sheet in the shape of frustum of a cone whose volume is
3310459
7cm . The radii of its lower and upper circular ends are 8cm and 20cm. find the
cost of metal sheet used in making container at rate of 21 4Rs per cm ?
Sol:
Given lower end radius of bucket 1 8r cm
Upper end radius of bucket
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
Class X Chapter 16 – Surface Areas and Volumes Maths
Let height of bucket be ‘h’
2 2 3
1
18 20 8 20
3V hcm ………(1)
Volume of milk container 3310459
4cm
3
2
73216
7V cm ……….(2)
Equating (1) and (2)
1 2V V
2 21 732168 20 8 20
3 7h
10459 42
653 45h
16h cm
Height of frustum cone 16h cm
Let slant height of frustum cone be ‘l’
2 2
1 2l r r h
2 220 8 16 144 256
20l cm
Slant height of frustum cone 20l cm
Total surface area of frustum cone
2 2
1 2 2 1r r l r r
2 2
20 8 20 20 8
560 400 64
2960 64 1024 3216 99cm
Total surface area 23216 99cm
Raman Choubay (raman.choubay@gmail.com)
Udaan School Of Mathematics
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